Number of Non Singular Matrices with Complex Elements

To find the number of distinct non-singular matrices in the set $S$, we first evaluate the determinant of the given matrix $A$:

$$A = \begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}$$

Expanding the determinant along the first row:

$$\det(A) = 1 \cdot (1 - c\omega) - a \cdot (\omega - c\omega^2) + b \cdot (\omega^2 - \omega^2)$$

Simplifying each term:

$$\det(A) = 1 - c\omega - a\omega + ac\omega^2 + b(0)$$ $$\det(A) = 1 - a\omega - c\omega + ac\omega^2$$

We can factorize this expression as:

$$\det(A) = (1 - a\omega)(1 - c\omega)$$

For the matrix $A$ to be non-singular, its determinant must be non-zero ($\det(A) \neq 0$). Thus, we require:

$$(1 - a\omega)(1 - c\omega) \neq 0 \implies 1 - a\omega \neq 0 \quad \text{and} \quad 1 - c\omega \neq 0$$

Since $a, b, c \in \{\omega, \omega^2\}$ and $\omega^3 = 1$ (with $\omega \neq 1$):

1. For $1 - a\omega \neq 0$:

   • If $a = \omega$, then $1 - a\omega = 1 - \omega^2 \neq 0$.
   • If $a = \omega^2$, then $1 - a\omega = 1 - \omega^3 = 1 - 1 = 0$.

   Therefore, we must choose $a = \omega$.

2. For $1 - c\omega \neq 0$:

   • Similarly, $c = \omega^2$ yields $1 - c\omega = 0$.

   Therefore, we must choose $c = \omega$.

3. For $b$:

   • Since $b$ is multiplied by $0$ in the determinant, the non-singularity condition is completely independent of $b$. Hence, $b$ can be either $\omega$ or $\omega^2$.

Thus, the possible ordered triples $(a, b, c)$ that yield a non-singular matrix are:

$$(\omega, \omega, \omega) \quad \text{and} \quad (\omega, \omega^2, \omega)$$

Therefore, the number of distinct non-singular matrices in the set $S$ is $2$.

Correct Option: A